gravity formula AC algebra 2 problem?

Question

the general formula for the height H after T seconds of a projectile launched upward form ground level with initial speed V is
h = vt -1/2gt^2 where g is the gravitational constant (approximatley 9.8 m/s^2 or 32 ft/s^2)

Show that the greatest height of the projectile is (v^2) / 2g.

Answer 1

v^2=u^2-2gh
at greatest height v=0
u^2=2gh
h=u^2/2g

aliter

Answer 2

This is an elementary Differential Calculus problem. First find the time when h is max, then substitute to find h-max.

1. Take the derivative of the height with respect to time and set equal to zero.

dh/dt = v - 2(1/2) gt
dh/dt = v-gt

set equal to zero:

0 = v-gt
t=v/g

So the projectile reaches it's highest point at time v/g

2. Now substitute into the height equation:

h = v(v/g) - 1/2 g(v/g)^2
h = v^2/g - gv^2/2g^2
h = v^2/g - v^2/2g
h = (2v^2-v^2)/2g
h = v^2/2g

QED

Answer 3

Rewrite h by completing the square.

First, rewrite it as h = -1/2g(t^2 - 2vt/g).

Now, use the fact that (t+a)^2 = t^2 + 2at + a^2 to rewrite t^2-(2v/g)t as (t-v/g)^2 - v^2/g^2. This can be seen by matching like terms t^2 + 2at + a^2 and t^2-(2v/g)t to obtain a=-v/g.

Put the -1/2g back, and you have

h = -1/2g(t-v/g)^2 + v^2/(2g).

(t-v/g)^2 >= 0 for all t, so -1/2g(t-v/g)^2 <= 0. Therefore, it is maximum when t=v/g. But then the whole term is zero, so you're left with the second term, v^2/(2g).

Answer 4

h=vt-(1/2)gt^2
h'=v-gt
h max when h'=0 so
v=gt ot
t=v/g
h max=v(v/g)-(1/2)g(v/g)^2=v^2/g-(1/2)v^2/g=v^2/(2g)